Integrand size = 11, antiderivative size = 92 \[ \int \frac {1}{\text {sech}^{\frac {3}{2}}(2 \log (c x))} \, dx=-\frac {3}{4 \left (c^4+\frac {1}{x^4}\right ) x^3 \text {sech}^{\frac {3}{2}}(2 \log (c x))}+\frac {x}{4 \text {sech}^{\frac {3}{2}}(2 \log (c x))}+\frac {3 \text {arctanh}\left (\sqrt {1+\frac {1}{c^4 x^4}}\right )}{4 c^4 \left (1+\frac {1}{c^4 x^4}\right )^{3/2} x^3 \text {sech}^{\frac {3}{2}}(2 \log (c x))} \] Output:
-3/4/(c^4+1/x^4)/x^3/sech(2*ln(c*x))^(3/2)+1/4*x/sech(2*ln(c*x))^(3/2)+3/4 *arctanh((1+1/c^4/x^4)^(1/2))/c^4/(1+1/c^4/x^4)^(3/2)/x^3/sech(2*ln(c*x))^ (3/2)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.08 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.70 \[ \int \frac {1}{\text {sech}^{\frac {3}{2}}(2 \log (c x))} \, dx=-\frac {\sqrt {1+c^4 x^4} \sqrt {\frac {c^2 x^2}{2+2 c^4 x^4}} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-\frac {1}{2},\frac {1}{2},-c^4 x^4\right )}{4 c^4 x^3} \] Input:
Integrate[Sech[2*Log[c*x]]^(-3/2),x]
Output:
-1/4*(Sqrt[1 + c^4*x^4]*Sqrt[(c^2*x^2)/(2 + 2*c^4*x^4)]*Hypergeometric2F1[ -3/2, -1/2, 1/2, -(c^4*x^4)])/(c^4*x^3)
Time = 0.26 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.636, Rules used = {6079, 6077, 798, 51, 60, 73, 220}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {1}{\text {sech}^{\frac {3}{2}}(2 \log (c x))} \, dx\) |
\(\Big \downarrow \) 6079 |
\(\displaystyle \frac {\int \frac {1}{\text {sech}^{\frac {3}{2}}(2 \log (c x))}d(c x)}{c}\) |
\(\Big \downarrow \) 6077 |
\(\displaystyle \frac {\int c^3 \left (1+\frac {1}{c^4 x^4}\right )^{3/2} x^3d(c x)}{c^4 x^3 \left (\frac {1}{c^4 x^4}+1\right )^{3/2} \text {sech}^{\frac {3}{2}}(2 \log (c x))}\) |
\(\Big \downarrow \) 798 |
\(\displaystyle -\frac {\int \frac {\left (1+\frac {1}{c^4 x^4}\right )^{3/2}}{c^2 x^2}d\frac {1}{c^4 x^4}}{4 c^4 x^3 \left (\frac {1}{c^4 x^4}+1\right )^{3/2} \text {sech}^{\frac {3}{2}}(2 \log (c x))}\) |
\(\Big \downarrow \) 51 |
\(\displaystyle -\frac {\frac {3}{2} \int \frac {\sqrt {1+\frac {1}{c^4 x^4}}}{c x}d\frac {1}{c^4 x^4}-\frac {\left (\frac {1}{c^4 x^4}+1\right )^{3/2}}{c x}}{4 c^4 x^3 \left (\frac {1}{c^4 x^4}+1\right )^{3/2} \text {sech}^{\frac {3}{2}}(2 \log (c x))}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle -\frac {\frac {3}{2} \left (\int \frac {1}{c \sqrt {1+\frac {1}{c^4 x^4}} x}d\frac {1}{c^4 x^4}+2 \sqrt {\frac {1}{c^4 x^4}+1}\right )-\frac {\left (\frac {1}{c^4 x^4}+1\right )^{3/2}}{c x}}{4 c^4 x^3 \left (\frac {1}{c^4 x^4}+1\right )^{3/2} \text {sech}^{\frac {3}{2}}(2 \log (c x))}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle -\frac {\frac {3}{2} \left (2 \int \frac {1}{c^2 x^2-1}d\sqrt {1+\frac {1}{c^4 x^4}}+2 \sqrt {\frac {1}{c^4 x^4}+1}\right )-\frac {\left (\frac {1}{c^4 x^4}+1\right )^{3/2}}{c x}}{4 c^4 x^3 \left (\frac {1}{c^4 x^4}+1\right )^{3/2} \text {sech}^{\frac {3}{2}}(2 \log (c x))}\) |
\(\Big \downarrow \) 220 |
\(\displaystyle -\frac {\frac {3}{2} \left (2 \sqrt {\frac {1}{c^4 x^4}+1}-2 \text {arctanh}\left (\sqrt {\frac {1}{c^4 x^4}+1}\right )\right )-\frac {\left (\frac {1}{c^4 x^4}+1\right )^{3/2}}{c x}}{4 c^4 x^3 \left (\frac {1}{c^4 x^4}+1\right )^{3/2} \text {sech}^{\frac {3}{2}}(2 \log (c x))}\) |
Input:
Int[Sech[2*Log[c*x]]^(-3/2),x]
Output:
-1/4*(-((1 + 1/(c^4*x^4))^(3/2)/(c*x)) + (3*(2*Sqrt[1 + 1/(c^4*x^4)] - 2*A rcTanh[Sqrt[1 + 1/(c^4*x^4)]]))/2)/(c^4*(1 + 1/(c^4*x^4))^(3/2)*x^3*Sech[2 *Log[c*x]]^(3/2))
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + 1))), x] - Simp[d*(n/(b*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n}, x ] && ILtQ[m, -1] && FractionQ[n] && GtQ[n, 0]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( b*(m + n + 1))) Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(- 1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[1/n Subst [Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]
Int[Sech[((a_.) + Log[x_]*(b_.))*(d_.)]^(p_.), x_Symbol] :> Simp[Sech[d*(a + b*Log[x])]^p*((1 + 1/(E^(2*a*d)*x^(2*b*d)))^p/x^((-b)*d*p)) Int[1/(x^(b *d*p)*(1 + 1/(E^(2*a*d)*x^(2*b*d)))^p), x], x] /; FreeQ[{a, b, d, p}, x] && !IntegerQ[p]
Int[Sech[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(d_.)]^(p_.), x_Symbol] :> S imp[x/(n*(c*x^n)^(1/n)) Subst[Int[x^(1/n - 1)*Sech[d*(a + b*Log[x])]^p, x ], x, c*x^n], x] /; FreeQ[{a, b, c, d, n, p}, x] && (NeQ[c, 1] || NeQ[n, 1] )
Time = 0.10 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.42
method | result | size |
risch | \(\frac {\left (c^{8} x^{8}-c^{4} x^{4}-2\right ) \sqrt {2}}{16 x \left (c^{4} x^{4}+1\right ) c^{2} \sqrt {\frac {c^{2} x^{2}}{c^{4} x^{4}+1}}}+\frac {3 c^{2} \ln \left (\frac {c^{4} x^{2}}{\sqrt {c^{4}}}+\sqrt {c^{4} x^{4}+1}\right ) \sqrt {2}\, x}{16 \sqrt {c^{4}}\, \sqrt {c^{4} x^{4}+1}\, \sqrt {\frac {c^{2} x^{2}}{c^{4} x^{4}+1}}}\) | \(131\) |
Input:
int(1/sech(2*ln(x*c))^(3/2),x,method=_RETURNVERBOSE)
Output:
1/16*(c^8*x^8-c^4*x^4-2)/x/(c^4*x^4+1)*2^(1/2)/c^2/(c^2*x^2/(c^4*x^4+1))^( 1/2)+3/16*c^2*ln(c^4*x^2/(c^4)^(1/2)+(c^4*x^4+1)^(1/2))/(c^4)^(1/2)*2^(1/2 )*x/(c^4*x^4+1)^(1/2)/(c^2*x^2/(c^4*x^4+1))^(1/2)
Time = 0.10 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.15 \[ \int \frac {1}{\text {sech}^{\frac {3}{2}}(2 \log (c x))} \, dx=\frac {3 \, \sqrt {2} c^{3} x^{3} \log \left (-2 \, c^{4} x^{4} - 2 \, {\left (c^{5} x^{5} + c x\right )} \sqrt {\frac {c^{2} x^{2}}{c^{4} x^{4} + 1}} - 1\right ) + 2 \, \sqrt {2} {\left (c^{8} x^{8} - c^{4} x^{4} - 2\right )} \sqrt {\frac {c^{2} x^{2}}{c^{4} x^{4} + 1}}}{32 \, c^{4} x^{3}} \] Input:
integrate(1/sech(2*log(c*x))^(3/2),x, algorithm="fricas")
Output:
1/32*(3*sqrt(2)*c^3*x^3*log(-2*c^4*x^4 - 2*(c^5*x^5 + c*x)*sqrt(c^2*x^2/(c ^4*x^4 + 1)) - 1) + 2*sqrt(2)*(c^8*x^8 - c^4*x^4 - 2)*sqrt(c^2*x^2/(c^4*x^ 4 + 1)))/(c^4*x^3)
\[ \int \frac {1}{\text {sech}^{\frac {3}{2}}(2 \log (c x))} \, dx=\int \frac {1}{\operatorname {sech}^{\frac {3}{2}}{\left (2 \log {\left (c x \right )} \right )}}\, dx \] Input:
integrate(1/sech(2*ln(c*x))**(3/2),x)
Output:
Integral(sech(2*log(c*x))**(-3/2), x)
\[ \int \frac {1}{\text {sech}^{\frac {3}{2}}(2 \log (c x))} \, dx=\int { \frac {1}{\operatorname {sech}\left (2 \, \log \left (c x\right )\right )^{\frac {3}{2}}} \,d x } \] Input:
integrate(1/sech(2*log(c*x))^(3/2),x, algorithm="maxima")
Output:
integrate(sech(2*log(c*x))^(-3/2), x)
Timed out. \[ \int \frac {1}{\text {sech}^{\frac {3}{2}}(2 \log (c x))} \, dx=\text {Timed out} \] Input:
integrate(1/sech(2*log(c*x))^(3/2),x, algorithm="giac")
Output:
Timed out
Timed out. \[ \int \frac {1}{\text {sech}^{\frac {3}{2}}(2 \log (c x))} \, dx=\int \frac {1}{{\left (\frac {1}{\mathrm {cosh}\left (2\,\ln \left (c\,x\right )\right )}\right )}^{3/2}} \,d x \] Input:
int(1/(1/cosh(2*log(c*x)))^(3/2),x)
Output:
int(1/(1/cosh(2*log(c*x)))^(3/2), x)
\[ \int \frac {1}{\text {sech}^{\frac {3}{2}}(2 \log (c x))} \, dx=\int \frac {\sqrt {\mathrm {sech}\left (2 \,\mathrm {log}\left (c x \right )\right )}}{\mathrm {sech}\left (2 \,\mathrm {log}\left (c x \right )\right )^{2}}d x \] Input:
int(1/sech(2*log(c*x))^(3/2),x)
Output:
int(sqrt(sech(2*log(c*x)))/sech(2*log(c*x))**2,x)